# AP Physics 1: Simple Harmonic Motion (SHM) - The Night Before 🚀 Hey! Let's get you ready to ace this exam! We're going to break down Simple Harmonic Motion (SHM) into bite-sized pieces. Remember, you've got this! 💪 ## Introduction to SHM SHM is all about things that oscillate or repeat their motion. Think of a swing, a spring, or a pendulum. We'll focus on frequency and period, which are key to understanding these movements. Let's dive in! ### Frequency and Period Basics * **Period (T):** The time it takes for one complete cycle or oscillation. Measured in seconds (s). 🔄 * **Frequency (f):** The number of cycles or oscillations per second. Measured in Hertz (Hz), where 1 Hz = 1 s⁻¹. * **Inverse Relationship:** Frequency and period are inversely related. As one goes up, the other goes down. Mathematically: $$f = \frac{1}{T}$$ or $$T = \frac{1}{f}$$ * **Example:** If a system has a frequency of 4 Hz, its period is 0.25 s. Think of it like this: **Frequency** is how *frequently* something happens, and **Period** is the *time* it takes to happen once. They're two sides of the same coin! 🪙 ## Object-Spring Oscillators ### Period of a Spring-Mass System * An object-spring oscillator involves a mass (m) attached to a spring, oscillating back and forth. * The **spring constant (k)** measures the stiffness of the spring. Higher k means a stiffer spring. * The period (T) of this system is determined by both the mass (m) and spring constant (k). The formula is: $$T = 2\pi \sqrt{\frac{m}{k}}$$ * **Mass (m) and Period (T):** Direct relationship. More mass = longer period. * **Spring Constant (k) and Period (T):** Inverse relationship. Stiffer spring = shorter period. * **Example:** If you double the mass, the period increases by a factor of √2. If you quadruple the spring constant, the period is halved. Remember, mass is directly under the square root in the numerator, while the spring constant is in the denominator. This helps you quickly see how changing them affects the period. ![Object-spring oscillator](https://upload.wikimedia.org/wikipedia/commons/thumb/7/79/Masse-Feder-System.svg/600px-Masse-Feder-System.svg.png)
*Caption: A mass-spring system undergoing simple harmonic motion.* ## Simple Pendulums ### Period of a Simple Pendulum * A simple pendulum consists of a mass (bob) hanging from a string of length (L), swinging back and forth. 🪀 * The period (T) depends on the length (L) of the pendulum and the acceleration due to gravity (g), approximately 9.8 m/s² on Earth. * The formula for the period of a simple pendulum is: $$T = 2\pi \sqrt{\frac{L}{g}}$$ * **Length (L) and Period (T):** Direct relationship. Longer pendulum = longer period. * **Gravity (g) and Period (T):** Inverse relationship. Higher gravity = shorter period. * **Mass of the bob:** Does *not* affect the period. 🤯 * **Example:** Doubling the length of the pendulum increases the period by a factor of √2. Don't confuse the formulas for spring-mass systems and pendulums! Mass is in the numerator for springs, but length is in the numerator for pendulums. Gravity (g) is only in the pendulum formula. ![Simple pendulum](https://upload.wikimedia.org/wikipedia/commons/thumb/9/9c/Simple_pendulum_diagram.svg/600px-Simple_pendulum_diagram.svg.png)
*Caption: A simple pendulum with length L and bob mass m.* ## Final Exam Focus Okay, let's nail down what's most important for the exam: * **Key Formulas:** Know $$f = \frac{1}{T}$$, $$T = 2\pi \sqrt{\frac{m}{k}}$$, and $$T = 2\pi \sqrt{\frac{L}{g}}$$ like the back of your hand. 📝 * **Relationships:** Understand how changing mass, spring constant, length, and gravity affects the period and frequency. * **Conceptual Questions:** Be ready to explain why the mass doesn't affect the period of a pendulum but does affect the period of a spring-mass system. * **Combining Concepts:** AP questions often mix SHM with energy and forces. Be prepared to apply these concepts in different scenarios. ### Last-Minute Tips * **Time Management:** Don't spend too long on one question. If you're stuck, move on and come back later. * **Units:** Always include units in your answers, and make sure they're correct. 📏 * **Show Your Work:** Even if you don't get the final answer, you can still get partial credit for showing your steps. * **Stay Calm:** Take a deep breath. You've studied hard, and you're ready for this! 🧘 ## Practice Questions ### Multiple Choice Questions 1. A mass-spring system oscillates with a period T. If the mass is doubled and the spring constant is halved, what is the new period? (A) T/2 (B) T (C) 2T (D) 2√2 T 2. A simple pendulum has a period of 2 seconds on Earth. If the same pendulum were taken to a planet where the acceleration due to gravity is four times that of Earth, what would be its period? (A) 1 s (B) 2 s (C) 4 s (D) 8 s 3. Which of the following changes would increase the period of a simple pendulum? (A) Increasing the mass of the bob (B) Decreasing the length of the pendulum (C) Increasing the length of the pendulum (D) Increasing the acceleration due to gravity ### Free Response Question A block of mass m = 0.5 kg is attached to a spring with spring constant k = 200 N/m on a frictionless horizontal surface. The block is pulled a distance of 0.1 m from its equilibrium position and released from rest. (a) Calculate the period of oscillation for the block-spring system. (2 points) (b) Calculate the maximum speed of the block during its oscillation. (3 points) (c) If the same block is now used as the bob of a simple pendulum with a length of 1 meter, what is the period of the pendulum's oscillation? (2 points) (d) Describe one change you could make to the pendulum to increase its period and explain why this change would have the desired effect. (2 points) ### FRQ Scoring Breakdown (a) Calculate the period of oscillation for the block-spring system. (2 points) * 1 point for using the correct formula: $$T = 2\pi \sqrt{\frac{m}{k}}$$ * 1 point for correct calculation: $$T = 2\pi \sqrt{\frac{0.5}{200}} = 0.314 s$$ (b) Calculate the maximum speed of the block during its oscillation. (3 points) * 1 point for recognizing that max speed occurs at equilibrium * 1 point for using the energy conservation approach: $$ \frac{1}{2} kA^{2} = \frac{1}{2} m v_{max}^{2}$$ * 1 point for correct calculation: $$v_{max} = A\sqrt{\frac{k}{m}} = 0.1 \sqrt{\frac{200}{0.5}} = 2 m/s$$ (c) If the same block is now used as the bob of a simple pendulum with a length of 1 meter, what is the period of the pendulum's oscillation? (2 points) * 1 point for using the correct formula: $$T = 2\pi \sqrt{\frac{L}{g}}$$ * 1 point for correct calculation: $$T = 2\pi \sqrt{\frac{1}{9.8}} = 2.01 s$$ (d) Describe one change you could make to the pendulum to increase its period and explain why this change would have the desired effect. (2 points) * 1 point for stating that increasing the length of the pendulum would increase the period * 1 point for explaining that period is proportional to the square root of the length: $$T \propto \sqrt{L}$$

Frequency and Period of SHM

Isabella Lopez

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